Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
(a) How does it provide magnification?
It is true that the angle subtended by the virtual image and the object at the eye is the same if the object is at the same position. However, the magnifying glass allows us to bring the object much closer to the eye than the normal Near Point (D = 25 cm).
The real benefit:
- Without the lens, you cannot focus on an object closer than 25 cm.
- With the lens, you can place the object closer (e.g., at 5 cm). This closer proximity increases the visual angle significantly compared to holding the object at 25 cm.
- Angular Magnification is the ratio of the angle subtended by the image (with lens) to the angle subtended by the object at the near point (without lens).
(b) Does magnification change if the eye is moved back?
Yes, it changes slightly.
If the eye is moved back from the lens, the angle subtended by the virtual image at the eye decreases. Therefore, the angular magnification slightly decreases.
Note: If the image is formed at infinity, the angle remains constant regardless of eye position, so magnification would not change in that specific case.
(c) Limitations of small focal length
While the formula (M = 1 + D/f) suggests that reducing focal length (f) increases power, there are practical limits:
- Aberrations: Lenses with very short focal lengths require very high curvature. This introduces severe Spherical and Chromatic Aberrations, making the image distorted and blurry.
- Grinding Difficulties: It is mechanically very difficult to grind and polish lenses with extremely small radii of curvature accurately.
(d) Why short focal lengths?
The total magnification (M) of a compound microscope is the product of the magnification of the objective (mo) and the eyepiece (me).
M = mo × me
- Objective: A short focal length (fo) is needed to produce a high linear magnification (since mo ≈ L / fo).
- Eyepiece: A short focal length (fe) is needed to act as a strong magnifying glass for the intermediate image (since me = 1 + D/fe).
Therefore, both lenses must have short focal lengths to maximize the total magnifying power.
(e) Position of the Eye (Exit Pupil)
Reason: The light rays emerging from the eyepiece converge at a specific small circular area called the Exit Pupil (or eye-ring). By positioning the eye at this exact point, all the light rays from the field of view enter the pupil, ensuring the brightest and widest possible view.
If the eye is placed too close (touching the lens), the iris blocks the peripheral rays, significantly reducing the field of view.
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